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Friends, we shall continue our discussion
on equal area criteria of stability.
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Today, we will cover I will address the following
problems. The basic definitions of a transient
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stability limit, the meaning of critical clearing
angle and critical clearing time then we shall
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study the effect of fault clearing time on
transient stability limit, next we shall study
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the effect of type of fault on stability when
we talk about the type of fault on stability
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we, I shall introduce the concept of fault
shunts then we will study the effect of grounding
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on stability.
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Now, before I address these issues I would
like to mention that this subject has been
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very well addressed in power system stability
written by Edward Wilson Kimbark there are
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3 volumes, volume 1, volume 2 and third volume
is now known as synchronous machines. These
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basic concepts are very well addressed in
volume 1.
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Now, let us define what the transient stability
limit. We have defined earlier the transient
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stability, now today we will emphasize on
this word transient stability limit. The transient
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stability limit refers to the amount of power
that can be transmitted through some point
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in the system with stability when the system
is subjected to severe aperiodic disturbance
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that here the stability limit is referred
in terms of the amount of power that is what
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is the power in mega watts that can be transmitted
and this is referred to some point in the
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system.
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Now the emphasis here is on some point in
the system. Now if you take a machine infinite
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bus system then there is only one transmission
line and therefore whatsoever power that can
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be transmitted on that line with stability
becomes the transient stability limit. However
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when I consider a multi machine system okay,
in that system this limit refers to a point
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in the power system that is there may be number
of transmission lines and you may refer that
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in this particular transmission line how much
power can be transmitted for a given operating
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condition with given type of severe disturbance,
given type of severe disturbance here.
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Now here we emphasize the word aperiodic the
disturbance is disturbance comes not in periodical
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manner but it comes in aperiodic manner it
comes and goes it is not actually that after
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every 5 second the disturbance keeps on coming.
Okay, this is what is the meaning of transient
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stability limit therefore, any power system
when we operate, we have to operate below
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transient stability limit.
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Okay so that it will withstand the the particular
type of disturbance for which the system is
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designed. Now next term is the critical clearing
angle. This critical clearing angle is specifically
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referred to a machine infinite bus system
because in the multi-machine system, you will
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have number of angles right and therefore
the definition to a multi-machine system is
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not that easily available.
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Now for a given system and for a given initial
load, there is a critical clearing angle,
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if the actual clearing angle is smaller than
the critical value a system is stable and
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if larger the system is unstable. I will explain
this point in detail okay in our further discussion
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that here the meaning is that there is some
critical clearing angle and actual clearing
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angle, if it is less than this value system
is stable in case the actual clearing angle
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is more than the critical clearing angle system
will become unstable.
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Similarly, we define another term critical
fault clearing time again for a given system
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and for given initial loading there is a critical
fault clearing time, if actual fault clearing
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time is smaller than the critical value the
system is stable, if larger then the system
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is unstable. Now this definition is applicable
to machine infinite bus system or a multi-machine
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system because whenever a system is operating
okay and if fault occurs on a particular element
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of the system then this fault is cleared by
removing the faulted element by operating
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the circuit breaker at the two ends and therefore
there is certain time required to clear this
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fault.
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In case this actual fault clearing time is
less than the critical clearing time system
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is stable otherwise, it is unstable further
if suppose the critical clearing time for
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a system comes out to be say .2 second and
actual fault clearing time is say .1 second
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then the difference .1 second is called the
stability margin. Okay therefore, we have
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been discussing in these days in terms of
what is the stability margin and stability
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margin can be quantified in terms of difference
in the critical clearing time for the system
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and actual fault clearing time because actual
fault clearing time depends upon the operating
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time of the protection system and circuit
breaker fault interruption time.
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Now let us understand what is the effect of
fault clearing time on transient stability
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limit, now here when I say it is a transient
stability limit it means it is a certain amount
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of power that can be transmitted without loss
of stability. The transient stability limit
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depends on type of fault and the duration
of fault, a very we shall state these aspects
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type of fault and duration of fault. The power
limit can be determined as a function of clearing
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angle suppose it is a machine infinite bus
system we can find out the transient stability
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limit or power limit as a function of clearing
angle and clearing time can be found by solving
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the swing equation up to the time of fault
clearing that if I want to know know the transient
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stability limit as a function of fault clearing
time.
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The approach we will discuss here but for
a machine infinite bus system the simple approach
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is that you first apply the equal area criteria,
find out for a given power what is the critical
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clearing angle and then once we know the angle,
we solve the swing equation up to the up to
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the fault clearing angle and corresponding
to that we read the value of time and that
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becomes our critical clearing time and therefore
when I say here, when I discussed earlier
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that when you apply this graphical method
that is equal area criteria of stability.
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We do not completely diverse the need for
solving swing equation but partially we do
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it wholly or partially this is what we are
the they are partially means partly you have
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to solve it and suppose a swing curve is required
to be solved for say 2 seconds for normal
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stability steady analysis but in this particular
case the time for which it is to be solved
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is very small suppose the for critical fault
clearing time comes out to be it is a .2 second
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okay, then I solve it from 0 to .2 second
not from 0 to 2 second and therefore this
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saves my time or computation time.
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As I have stated that fault clearing time
is sum of the time that the protective relays
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take to close the circuit breaker trip circuit
and the time required for the circuit breaker
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to interrupt the fault current.
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The general conclusion that decrease in fault
clearing time improves stability and increases
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transient stability limit is just as valid
for a multi machine system as for a 2 machine
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system. This point I stated earlier also again
we reiterate that general conclusion is the
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decrease in fault clearing time improves stability
and it improves the transient stability limit
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okay.
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Now this conclusion is valid for machine infinite
bus system 2 machine system and even for a
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multi machine system and that is why the efforts
have been made all through to reduce the fault
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clearing time this has been possible by applying
fast acting protective relays and fast acting
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circuit breakers.
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Now to illustrate this that how do we calculate
the transient stability limit and obtain a
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curve relating the transient stability limit
and the fault clearing time, we will consider
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these 2 machine system and a generator infinite
bus and consider that the fault occurred at
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the middle of the line. Now for this system
we can find out the 3 power angle curves,
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pre fault, during the fault and post fault,
once we know this power angle curves we can
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apply the equal area criteria and determine
the certain points on the transient stability
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limit versus the fault clearing time.
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Now we will consider the 2 extreme cases,
first case we will consider that the fault
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is instantaneously clear. Suppose fault occurs
in the system and the time which it takes
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to clear the fault is instantaneous it does
not have any time practically it does not
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happen okay. Now this can also be considered
similar to that one transmission line is stripped,
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okay by operating the circuit breakers.
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Now in that case if you want to find out what
is the transient stability limit then the
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approach would be you make use of these two
power angle characteristic, pre fault output
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characteristic and the post fault output where
during fault we do not require it because
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the system has not operated with fault on
the system.
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Now in this case what we will do is, this
mechanical input line, the mechanical input
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line Pm is moved up and down, Pm is moved
up and down till these two areas are equal
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that is a1 and a2 is the area bounded by mechanical
input line, the post fault power angle curve
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that is from delta naught to delta 1 and a2
is again the area bounded by the post fault
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power angle curve and the mechanical input
line but they have opposite signs.
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Now when these two areas are equal that will
give us the stability limit that is this Pm
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becomes the stability limit you have, what
a what is to be done is that you have to move
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this mechanical input line up and down you
have to do 1 or 2 you know iterations and
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the movement these two areas becomes equal
that becomes the stability limit because here
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the fault has been cleared instantaneously.
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Now we will take another extreme case, where
the the sustained fault on the system, that
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is fault is not cleared in this situation
we require the two power angle curves, one
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is the pre fault another is the during the
fault power angle curve. Now these 2 curves
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are plotted here again the approach will be
that you move the mechanical input line up
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and down till these 2 areas a1a2 are equal
that is when you do this computations you
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assume some value of Pm.
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You know the value of delta naught initial
operating angle, you can find out what is
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the value of this angle delta 1 that will
depend upon the intersection of mechanical
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input and fault on power angle curve. Okay
similarly, you can find out delta m which
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is equal to phi minus delta 1, okay now you
find out this area by process of integration
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and you equate this with the area a2, in case
these 2 areas are equal then this Pm becomes
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the transient stability limit with sustained
fault. Now the third situation will be that
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the fault is clear infinite time.
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Now under this situation we require all the
3 power angle curves, okay. Now one way is
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that you move this mechanical input line Pm
up and down and see that these 2 areas are
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equal and but this there actually we required
the information and what is the fault clearing
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angle there were 2 parameters involved, one
is the fault clearing angle another is the
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mechanical input line. The easiest process
will be we assume some value of Pm and instead
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of moving this mechanical input line up and
down you adjust this clearing angle, you will
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you move this line either on left or on right
it means you assume some value of fault clearing
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angle and see whether these 2 areas are equal.
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In case you find actually that for a assumed
value of angle a1 is greater than a2 move
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this line on left so that a1 decreases and
a2 increases right and you do this exercise
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still these 2 areas are equal it means what
we have done now here is for given input Pm
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we have obtained the
critical clearing angle and then we integrate
the swing equation from this point delta naught
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up to this critical clearing angle and find
out the corresponding value of critical clearing
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time.
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Now this is the way you can find out a number
of points assume some value, suppose the value
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of Pm is small you will find that delta c
will be very large and a stage may come when
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delta c will coincide with delta m right that
is the case for sustained fault that is when
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you are moving here right and if this fault
you find actually that Pm is such that system
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is stable when delta c equal to delta m that
is the condition for sustained fault then
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when you, if you are moving if the Pm is moved
up and when you are you have to move this
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line the moment you find actually the delta
critical clearing angle delta c same as delta
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naught that become the instantaneous occurring
time and therefore by this approach we can
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plot the curve relating the transient stability
limit versus the fault clearing time in seconds.
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For example, this graph shows for a typical
system on this x axis by plotting the fault
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clearing time in seconds, y axis we are plotting
the stability limit in per unit and the stability
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limit of the system when the fault is cleared
instantaneously is denoted as 1 per unit and
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for all other fault clearing time it is going
to be less than this okay.
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Now here actually this .2 shows the sustained
condition because there is a break here in
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the graph right because when the fault is
sustained the stability limit is going to
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very small, sustained fault condition there
is a fault is there and system is not losing
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stability it means that is the situation which
occurs only when the fault, I am sorry a a
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the power output is small right. Further as
we have seen that the post fault, I am sorry
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the post fault power angle characteristic
depends upon the element which has been removed
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and the remaining system.
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Similarly, the during the fault power angle
curve is concerned it depends upon the location
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of fault on the transmission line. Suppose
you consider the 2 machine system and for
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the whole analysis what we have done is we
have assumed the fault in the middle of the
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line suppose I shift the fault location from
the middle toward the sending end of the line
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or you shift this from middle to the receiving
end of the line in that case you will find
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that the stability limit will be different
the curve relating the stability limit versus
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the fault clearing time will be different.
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Now these 2 curves show the fault at sending
end of the line and this show the fault at
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middle of the line now you can easily see
that where the fault occurs at the sending
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end of the line okay what will happen to be
power angle curve the during the fault condition
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during the fault condition no power will be
can be transmitted on this
when you consider the machine infinite bus
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system let us say the system right, if suppose
the fault occurs right at the sending end
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of the line then this fault is as good as
a fault on this bus.
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Okay and therefore the voltage at this bus
becomes 0 or it collapses because I am considering
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here a balanced 3-phase fault and therefore
no power can be transmitted from generator
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to the infinite bus right, therefore you will
find actually the r2 r2 which is multiplying
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r2 Pmax sin delta r2 will become 0. This is
a very special case and you will find that
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the transient stability limit will be less
as compared to where the fault occurs in any
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other point on the transmission line that
is why these 2 curves have been plotted to
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illustrate the effect of location of fault
on the transmission line.
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Now
by applying the equal area criteria of stability,
we can find out for a given value of mechanical
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input and for given fault location we can
compute the value of the angle fault clearing
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angle delta at which the system just stable
that is we can compute the critical clearing
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angle for computing the critical clearing
angle what is to be done is that you find
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out this area a1 that is you integrate, you
integrate that is a1 can be written as integral
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00:26:01,789 --> 00:26:08,789
of delta naught to delta c Pm mechanical input
minus Pe1 not P1 Pe2 d delta where Pe2 is
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00:26:25,080 --> 00:26:32,080
equal to r2 times Pmax sin delta okay.
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00:26:44,450 --> 00:26:51,450
Then
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00:27:00,200 --> 00:27:07,200
the area a2 can be computed integral delta
c to delta M, here we will be writing this
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00:27:21,809 --> 00:27:28,809
as Pe3 minus Pm d delta where Pe3 is equal
to r2 Pmax sin. Okay, if you equate these
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2 areas that is you equate a1 with a2 and
you can find the expression for critical clearing
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00:28:03,470 --> 00:28:10,470
angle or the equation for computing critical
clearing angle. This equation has been obtained
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00:28:14,100 --> 00:28:21,100
and it is a cos delta c equal to delta m minus
delta o, sin delta o minus r1 cos delta o
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00:28:22,299 --> 00:28:29,299
plus r2 cos delta m divided by r2 minus r1
this, this expression can be derived without
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00:28:30,919 --> 00:28:36,019
any difficulty by equating those 2 expressions.
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Now, when you apply this formula, you have
ensure that these angles are the delta m delta
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naught they are they are yes substituted in
radians sometimes people committed mistake
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they put directly in degrees and therefore
the result will be observed.
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Now this angle delta m has to be calculated
by using this formula delta m is equal to
204
00:29:14,130 --> 00:29:21,130
phi minus sin inverse Pm divided by r2Pmax
because delta m is delta m is obtained where
205
00:29:26,360 --> 00:29:33,360
the where the post fault power angle characteristic
intersects with mechanical input line like
206
00:29:37,309 --> 00:29:44,309
it intersect at 2 points one is the angle
which is given by this this equation and another
207
00:29:45,490 --> 00:29:52,490
will be phi minus this, therefore delta m
is equal to phi minus this therefore another
208
00:29:52,730 --> 00:29:58,570
sometime people commit mistake in computing
the value of delta m correctly and where this
209
00:29:58,570 --> 00:30:05,570
r1 is the x12 before fault and x12 during
fault r2 is x12 before fault x12 after fault,
210
00:30:08,509 --> 00:30:15,269
that is x12 is the reactance connecting the
nodes that is internal voltage of the generator
211
00:30:15,269 --> 00:30:22,269
to the infinite bus voltage and this formula
is applicable to a 2 machine system only we
212
00:30:22,399 --> 00:30:29,399
do not have such formula for a multi machine
system.
213
00:30:33,259 --> 00:30:39,710
Now next point we have to address is how to
determine power angle curve for unsymmetrical
214
00:30:39,710 --> 00:30:46,710
fault, till now till now we have assumed a
balanced 3-phase fault for our analysis and
215
00:30:50,330 --> 00:30:57,330
we also assume that this 3-phase fault is
a metallic short circuit there is no fault
216
00:30:58,159 --> 00:31:05,159
impedance involved, under this situation the,
the faulted point is directly connected to
217
00:31:10,399 --> 00:31:17,399
the reference bus in the equivalent network
and we analyze it but the moment you have
218
00:31:18,870 --> 00:31:25,870
unbalanced fault then things cannot be as
simple as in a 3-phase system because as you
219
00:31:26,309 --> 00:31:32,639
know actually that unbalanced faults can be
analyzed by by using the method of symmetrical
220
00:31:32,639 --> 00:31:36,899
components.
221
00:31:36,899 --> 00:31:43,289
Okay and when we apply the method of symmetrical
components we will come across positive sequence
222
00:31:43,289 --> 00:31:50,289
network negative sequence network zero sequence
network and we can compute depending upon
223
00:31:50,840 --> 00:31:55,870
the type of fault the positive sequence currents,
negative sequence currents, zero sequence
224
00:31:55,870 --> 00:31:56,379
currents.
225
00:31:56,379 --> 00:32:03,379
We also know that how to draw for a given
system the positive sequence network, negative
226
00:32:06,649 --> 00:32:13,649
sequence network and zero sequence networks
before I tell you how to account for the unsymmetrical
227
00:32:16,250 --> 00:32:23,250
fault for determining power angle curve during
fault condition. We have to understand some
228
00:32:28,000 --> 00:32:35,000
basic concepts one basic concept is which
I will introduce and explain in subsequent
229
00:32:37,309 --> 00:32:40,769
discussion.
230
00:32:40,769 --> 00:32:46,960
The concept of fault shunt now here to ah
before we understand this fault shunt, let
231
00:32:46,960 --> 00:32:51,980
us understand a since the internal electromotive
forces of 3-phase synchronous machine are
232
00:32:51,980 --> 00:32:57,179
of positive sequence that is so for the three
phase synchronous generators are concerned
233
00:32:57,179 --> 00:33:03,919
we always generate positive sequence voltages.
We do not generate negative sequence or zero
234
00:33:03,919 --> 00:33:10,919
sequence voltages, no power results from interaction
of positive sequence voltages with negative
235
00:33:12,509 --> 00:33:19,509
or zero sequence currents. Although, the stator
may be carrying negative sequence current,
236
00:33:21,509 --> 00:33:25,169
zero sequence current but when this negative
sequence current interacts with the positive
237
00:33:25,169 --> 00:33:28,970
sequence voltages no power is generated no
average power is generated.
238
00:33:28,970 --> 00:33:34,090
Similarly, no average power is generated when
positive sequence voltages interact with 0
239
00:33:34,090 --> 00:33:41,090
sequence current, okay and therefore to compute
the power angle characteristic which basically
240
00:33:43,570 --> 00:33:50,570
relates with the power transfer from machine
to the infinite bus okay. We are we have to,
241
00:33:53,840 --> 00:34:00,600
we have to compute primarily the positive
sequence currents okay and to compute the
242
00:34:00,600 --> 00:34:07,600
positive sequence currents as we know actually
that during fault condition the positive sequence,
243
00:34:07,679 --> 00:34:13,280
negative sequence zero sequence networks are
connected in a particular fashion if suppose
244
00:34:13,280 --> 00:34:18,250
there is a line to ground fault these three
networks will be connected in series if it
245
00:34:18,250 --> 00:34:23,850
is a line to line fault then these two networks
will be connected in parallel looking into
246
00:34:23,850 --> 00:34:28,890
the looking into the faulted terminals.
247
00:34:28,890 --> 00:34:34,380
We have to look where do we connect in parallel
where the fault occurs in two faulted points
248
00:34:34,380 --> 00:34:38,880
similarly, we have it is a double line to
ground fault then the three networks will
249
00:34:38,880 --> 00:34:45,520
be connected in parallel.
250
00:34:45,520 --> 00:34:52,520
The simplest approach to account for the unbalanced
or unsymmetrical fault is by connecting shunt
251
00:34:53,840 --> 00:35:00,840
impedance ZF at the point of fault that is
in the positive sequence network we return
252
00:35:02,070 --> 00:35:07,160
the positive sequence network as it is earlier
between the fault point and the reference
253
00:35:07,160 --> 00:35:11,090
we were connecting zero impedance that is
directly connected.
254
00:35:11,090 --> 00:35:16,940
However, when unbalanced or unsymmetrical
fault is there we have to connect an impedance
255
00:35:16,940 --> 00:35:23,940
of value ZF that is called fault shunt ZF.
The value of ZF depends upon the impedance
256
00:35:28,950 --> 00:35:35,950
Z2 and Zo of the negative and zero sequence
networks viewed from the point of fault that
257
00:35:45,390 --> 00:35:52,190
is ZF is function of positive sequence I am
sorry, negative sequence and zero sequence
258
00:35:52,190 --> 00:35:59,190
impedances. Here, I will without a derivation
right now I am just giving the result the
259
00:36:06,770 --> 00:36:13,770
this table shows the type of short circuit
and the fault shunt ZF is the impedance of
260
00:36:14,780 --> 00:36:21,780
the fault shunt, if it is a line to ground
fault the value of ZF is Z0 plus Z2, if it
261
00:36:22,920 --> 00:36:29,920
is a line to line fault the fault shunt impedance
is Z2, if it is a double line to ground fault
262
00:36:32,740 --> 00:36:39,740
if the fault shunt is the parallel combination
of Z0 and Z2 and if it is 3 phase fault ZF
263
00:36:43,490 --> 00:36:47,590
is 0 okay.
264
00:36:47,590 --> 00:36:52,670
Now this table is very important and I will
show you a list through illustration, how
265
00:36:52,670 --> 00:36:59,670
do we get this in time, now a typical statistics
of the occurrence of type of faults or frequency
266
00:37:05,180 --> 00:37:07,460
of occurrence type of fault.
267
00:37:07,460 --> 00:37:14,460
A typical 132 K. V system the data were obtained
and out of the total 72 faults which occurred
268
00:37:19,060 --> 00:37:26,060
on the system, 58 were lined to ground faults
double line to ground faults were 8 and 3-phase
269
00:37:26,730 --> 00:37:33,160
faults were 6 in fact line to line faults
generally gets converted into line to line
270
00:37:33,160 --> 00:37:40,160
to ground fault. You can see very easily here
that the frequency of occurrence of 3- phase
271
00:37:43,160 --> 00:37:50,160
fault is lowest and the the frequency of occurrence
of line to ground fault is the highest and
272
00:37:54,480 --> 00:38:01,210
therefore, in case you design your system
or operating condition of the system is design,
273
00:38:01,210 --> 00:38:08,210
considering 3- phase fault it means we are
very very pessimistic in our approach, it
274
00:38:10,380 --> 00:38:15,920
may be assumed actually that faults are would
be very severe and we are taking a very safe
275
00:38:15,920 --> 00:38:16,500
margin.
276
00:38:16,500 --> 00:38:20,930
However, if you apply only considering line
to ground fault definitely you are optimistic
277
00:38:20,930 --> 00:38:26,290
where you feel that these faults may not occur
because if you design considering line to
278
00:38:26,290 --> 00:38:33,050
ground fault and if 3- phase fault occurs,
system is going to lose stability similarly
279
00:38:33,050 --> 00:38:36,500
double line to ground fault occurs it is going
to loose stability in case you do not have
280
00:38:36,500 --> 00:38:41,160
any margin right and therefore the practices
I will tell you what are the practices which
281
00:38:41,160 --> 00:38:46,380
are as followed ah for designing the system
because we have to make a balance.
282
00:38:46,380 --> 00:38:53,380
We have, we should not be very optimistic
we should not be very pessimistic in our approach
283
00:38:57,830 --> 00:39:04,830
before I tell you about the effect of grounding,
let us just see how do we account and compute
284
00:39:13,040 --> 00:39:20,040
the value of fault shunt impedance ZF okay.
I have told you actually that fault shunt
285
00:39:21,490 --> 00:39:28,380
impedance ZF is different for different fault
and the the table had been shown to show you
286
00:39:28,380 --> 00:39:30,300
the expressions.
287
00:39:30,300 --> 00:39:36,170
Okay let us take this the simple machine infinite
bus problem where you have generator neutral
288
00:39:36,170 --> 00:39:43,170
of the generate is grounded double circuit
transmission line. In this case I have taken
289
00:39:44,870 --> 00:39:49,930
only one transformer there is no transformer
shown here but if that is there is a transformer
290
00:39:49,930 --> 00:39:55,980
here that can also be considered this delta
star connected transformer start point grounded
291
00:39:55,980 --> 00:39:59,550
infinite bus system.
292
00:39:59,550 --> 00:40:06,550
Now when you solve the problem considering
unsymmetrical fault we need information about
293
00:40:07,070 --> 00:40:14,070
positive negative and zero sequence reactance’s
of the system components for generator Xd
294
00:40:14,170 --> 00:40:21,170
prime is 0.35, the negative sequence reactance
of the generator is less than Xd prime is
295
00:40:23,310 --> 00:40:29,330
0.24, X naught is the lowest that is zero
sequence reactance is always low 0.06, for
296
00:40:29,330 --> 00:40:34,570
the transmission line is concerned its positive
and negative sequence reactance’s are equal
297
00:40:34,570 --> 00:40:41,570
that is Z 1 is 0.4 and j times 0.4 the and
the negative sequence reactance is also 0.4
298
00:40:46,530 --> 00:40:53,530
the zero sequence reactance of the transmission
line is always more than the positive or negative
299
00:40:55,300 --> 00:40:59,830
sequence reactance, in this case it is 0.65
the typical values it may be even more it
300
00:40:59,830 --> 00:41:05,020
may be sometimes 2 to 2.5 times even 2 to
3 three times it all depends upon this system.
301
00:41:05,020 --> 00:41:12,020
For a transformer we can assume this X1 X2
X0 equal to .1 that is they are equal, with
302
00:41:13,250 --> 00:41:20,250
this
the data’s which we have assumed let us
first obtain the positive sequence, negative
303
00:41:28,560 --> 00:41:31,230
sequence and zero sequence networks.
304
00:41:31,230 --> 00:41:38,230
The positive sequence network
is simple is same as what you do actually
for analyzing for balance 3-phase fault condition,
305
00:41:43,960 --> 00:41:50,960
this is the internal voltage E prime the direct
axis transient reactance I am putting the
306
00:41:57,950 --> 00:42:04,950
reactance value only 0.35 transmission line
reactance 0.4, 0.4 and the transformer reactance
307
00:42:24,120 --> 00:42:31,120
here 0.1 and the infinite bus voltage okay
this is the voltage V. I am showing this as
308
00:42:42,550 --> 00:42:49,550
a voltage V, now V we shall consider that
the fault occurs at the sending end of one
309
00:42:49,890 --> 00:42:56,610
of the transmission lines right at this point.
This is the sending end of the line and therefore
310
00:42:56,610 --> 00:43:02,730
I will added this is as good as the fault
occurring at the bus therefore let us call
311
00:43:02,730 --> 00:43:09,730
this point as the point P on the series and
this is our reference.
312
00:43:12,410 --> 00:43:19,410
Now so far the negative sequence network is
concerned it is it will have the same structure
313
00:43:27,190 --> 00:43:31,080
except that the sources will not be present
there are no sources because we do not generate
314
00:43:31,080 --> 00:43:38,080
the negative sequence voltages and therefore
the negative sequence network can be drawn.
315
00:43:55,190 --> 00:44:02,190
This is my reference bus, okay we call this
o this point continues to be P now what we
316
00:44:11,120 --> 00:44:18,120
will do here is that this point we will call
as P2 that is the fault point in the negative
317
00:44:22,170 --> 00:44:29,170
sequence network and we will call this point
as P1 in the positive sequence network, the
318
00:44:30,070 --> 00:44:37,070
points are same P1 and P2 are same on the
physical system. The values here are now because
319
00:44:40,230 --> 00:44:47,230
in a zero sequence, a negative sequence network,
the generator reactance is 0.24 per unit,
320
00:44:47,660 --> 00:44:54,660
the transmission line is same 0.4, 0.4 transformer
is also 0.1 okay.
321
00:44:58,910 --> 00:45:05,440
Now we can find out the equivalent reactance
of the negative sequence network looking into
322
00:45:05,440 --> 00:45:12,440
this points P2 and o, this exercise when you
do you will find actually that equivalent
323
00:45:13,540 --> 00:45:20,540
comes out to be a reactance whose value is
0.133 for this problem you can say this is
324
00:45:23,510 --> 00:45:30,510
P2 and you can even call this as a o2 the
reference bus for the negative sequence network,
325
00:45:33,190 --> 00:45:40,190
therefore these 2 terminals are important
for us. The next step is to draw zero sequence
326
00:45:43,710 --> 00:45:48,440
I mean to draw the zero sequence network for
the system.
327
00:45:48,440 --> 00:45:55,440
When we draw the zero sequence network, we
have to consider the connection of transformers
328
00:45:57,750 --> 00:46:04,750
because transformers may be connected in different
modes and we have to also consider the neutral
329
00:46:05,520 --> 00:46:11,110
impedance. In case you have put a impedance,
in the neutral circuit in the equivalent circuit
330
00:46:11,110 --> 00:46:15,730
that impedance will appear as 3 times the
actual value in this particular case, the
331
00:46:15,730 --> 00:46:22,660
neutral have been solidly grounded therefore
they do not appear.
332
00:46:22,660 --> 00:46:29,660
In this particular case if you draw this zero
sequence network it will come out to be
this value is 0.65, this is also 0.65, this
333
00:46:44,300 --> 00:46:51,300
is 0.1 this is 0.06, this is the zero sequence
reactance of the generator. This is our reference
334
00:46:58,570 --> 00:47:05,570
bus incidentally in this connection, the generator
neutrally is grounded therefore we can connect
335
00:47:08,360 --> 00:47:14,620
a neutral point to the reference. Okay if
you had it not been grounded this would have
336
00:47:14,620 --> 00:47:21,620
been open the transformer is a start delta
transformer with start point grounded and
337
00:47:22,410 --> 00:47:27,030
therefore again this point will be connected.
338
00:47:27,030 --> 00:47:30,870
Now those who do not have the practice of
drawing the zero sequence network, I will
339
00:47:30,870 --> 00:47:37,870
advice them to refresh their knowledge about
drawing the positive negative in zero sequence
340
00:47:39,080 --> 00:47:46,080
networks particularly zero sequence networks,
considering the different types of transformer
341
00:47:46,210 --> 00:47:53,210
connections now here this is the fault point
I will call this as a P000 and the equivalent
342
00:47:55,800 --> 00:48:01,110
impedance looking into these two terminals
has been computed it comes out to be equal
343
00:48:01,110 --> 00:48:08,110
to 0.053 these points are Po and okay this
so far we have obtained the positive negative
344
00:48:18,890 --> 00:48:25,040
in zero sequence networks and we have also
obtained the equivalent value of the positive,
345
00:48:25,040 --> 00:48:32,040
negative sequence impedances for the network
considering the fault location.
346
00:48:32,910 --> 00:48:39,910
Now in order to consider the effect of different
type of fault in the system right, these 3
347
00:48:40,290 --> 00:48:47,280
networks have to be connected in a proper
fashion. They for line to ground fault which
348
00:48:47,280 --> 00:48:51,430
we have considered these 3 networks can be
connected in this fashion.
349
00:48:51,430 --> 00:48:58,430
This diagram shows the positive sequence network
and here we have not simplified this network,
350
00:48:58,770 --> 00:49:05,770
the fault occurs at the point P1 in the negative
sequence network is shown in the terminals
351
00:49:07,680 --> 00:49:14,680
P2 and O2, well while zero sequence impedance
is on between the terminals Po and Oo.
352
00:49:17,690 --> 00:49:24,690
Now for simulating line to ground fault the
3 networks positive, negative and zero sequence
353
00:49:26,420 --> 00:49:31,530
networks are connected in series that how
we can connect these 3 networks in series
354
00:49:31,530 --> 00:49:38,530
that is O1 is connected to P2, O2 is connected
to Po and Oo is connected to P1 this becomes
355
00:49:38,780 --> 00:49:45,780
a series connection. It can be very clearly
seen here that the positive sequence network
356
00:49:46,580 --> 00:49:53,580
is modified and here we have now and the impedance
connected between P1 and O1 for line to ground
357
00:49:58,590 --> 00:50:05,590
fault, the value of the impedance connected
is Z2 plus Z0 and this becomes the fault shunt.
358
00:50:10,140 --> 00:50:17,140
Now this network can be simplified, I will
put in this simple form here between the node
359
00:50:18,920 --> 00:50:25,920
1 and 3 the reactance is the transient direct
axis transient reactance of the generator,
360
00:50:27,750 --> 00:50:33,580
this is the equivalent reactance of transmission
line and the transformer and between this
361
00:50:33,580 --> 00:50:40,580
node 3 and O, reference node we have connected
the fault shunt. If we have considered the
362
00:50:41,790 --> 00:50:48,790
lossless system then the fault shunt impedance
becomes a pure reactance that is j times XF.
363
00:50:50,610 --> 00:50:57,610
This network can be further transformed that
is the start connected three impedances can
364
00:50:58,500 --> 00:51:05,500
be replaced by an equivalent delta connected
impedances.
365
00:51:07,950 --> 00:51:14,950
This figure shows the equivalent delta the
nodes 1 and 2 are retained, the reference
366
00:51:16,870 --> 00:51:23,870
node is also retained as it is now the reactance
connecting the node 1and 2 is X12 and the
367
00:51:26,700 --> 00:51:33,700
reactance which is directly coming across
the E prime source E prime is shown here.
368
00:51:35,490 --> 00:51:40,820
Similarly, the reactance coming directly across
the infinite bus voltage is also coming and
369
00:51:40,820 --> 00:51:46,440
is shown here now know that these 2 reactance’s
do not affect the power transfer capability
370
00:51:46,440 --> 00:51:52,960
or power transmission capability of the system
therefore we concentrate only on the reactance
371
00:51:52,960 --> 00:51:59,960
connecting the nodes 1and 2. Now for line
to ground fault X12 is equal to 0.35 plus
372
00:52:02,410 --> 00:52:09,410
0.3 plus 0.35 into 0.3 divided by XF where
XF will be the sum of the 2 impedances that
373
00:52:14,770 --> 00:52:20,310
is zero sequence and negative sequence impedance.
374
00:52:20,310 --> 00:52:27,160
Now the value of XF will be different for
different type of fault. The value of X12
375
00:52:27,160 --> 00:52:33,710
is computed for three different types of unsymmetrical
faults for line to ground fault condition
376
00:52:33,710 --> 00:52:40,710
the X 12 is 1.22 per unit, for line to line
fault X12 comes out to be 1.44 per unit and
377
00:52:41,160 --> 00:52:46,690
for double line to ground fault it comes out
to be 3.41 per unit.
378
00:52:46,690 --> 00:52:53,690
We can see here that the reactance connecting
the nodes 1 and 2 which is primarily affects
379
00:52:57,350 --> 00:53:04,350
the power flow on the transmission line increases
as we go from line to ground fault to double
380
00:53:04,970 --> 00:53:11,970
line to ground fault, if we determine the
power angle characteristic during fault considering
381
00:53:15,940 --> 00:53:22,740
different type of faults for line to ground
fault, for this particular system consider
382
00:53:22,740 --> 00:53:28,110
the power angle characteristic comes out to
be P2 equal to 0.82 sin delta, for line to
383
00:53:28,110 --> 00:53:35,110
line fault the power angle curve is P2 equal
to 0.695 sin delta and double line to ground
384
00:53:35,900 --> 00:53:42,900
fault P2 is 0.293 sin delta that is if you
examine these 3 power angle curves we find
385
00:53:43,460 --> 00:53:50,460
that the power angle curve with line to ground
fault has the highest amplitude while the
386
00:53:51,670 --> 00:53:56,230
power angle curve corresponding to line to
line to ground fault or double line to ground
387
00:53:56,230 --> 00:54:03,230
fault is having the smallest amplitude or
and hence the, from the consideration of the
388
00:54:09,760 --> 00:54:15,480
transient stability limit or the power which
can be transferred without loss of synchronism,
389
00:54:15,480 --> 00:54:22,480
the line to ground fault will provide more
transient stability limit as compared to double
390
00:54:23,600 --> 00:54:29,570
line to ground fault.
391
00:54:29,570 --> 00:54:36,570
This figure shows the plot of transient stability
limit or power limit as a function of fault
392
00:54:38,630 --> 00:54:45,630
duration in seconds. This curve shows the
relationship between transient stability limit
393
00:54:47,610 --> 00:54:54,230
and fault duration for line to ground fault,
the second curve is for line to line fault,
394
00:54:54,230 --> 00:55:00,650
third curve is for 2 line to ground fault
and the last curve is plotted for a 3- phase
395
00:55:00,650 --> 00:55:02,900
fault.
396
00:55:02,900 --> 00:55:09,560
Now we can easily see here that in case the
fault is clear instantaneously that is when
397
00:55:09,560 --> 00:55:16,560
the fault duration is 0, then the transient
stability limit is same in all the 4 cases
398
00:55:17,700 --> 00:55:24,700
and therefore we can conclude that the transient
stability limit is not affected by the type
399
00:55:24,810 --> 00:55:28,150
of fault, if the fault is cleared instantaneously.
400
00:55:28,150 --> 00:55:35,150
However, if the fault is cleared the time
delay then it is very clear actually that
401
00:55:36,630 --> 00:55:42,130
the transient stability limit is lowest when
3- phase fault occurs and transient stability
402
00:55:42,130 --> 00:55:47,220
limit is highest for line to ground fault
therefore, we can see that from the point
403
00:55:47,220 --> 00:55:54,220
of view of severity the line to ground fault
is the least severe as compared to 3- phase
404
00:55:54,360 --> 00:56:01,360
fault or we can say that 3- phase fault is
the severest fault from the stability consideration.
405
00:56:03,610 --> 00:56:08,420
Now we study the effect of grounding on stability
the methods of grounding of a power system
406
00:56:08,420 --> 00:56:14,950
modify the 0 sequence impedance this affects
the impedance of the fault shunts for representing
407
00:56:14,950 --> 00:56:21,520
the ground faults and thereby affect the severity
of such faults.
408
00:56:21,520 --> 00:56:27,570
A typical to a system has been examined and
transient stability limit is computed for
409
00:56:27,570 --> 00:56:34,570
different values of ZS and ZR for a 2 machine
system.
410
00:56:34,620 --> 00:56:41,620
Here the value of ZS and ZR are varied from
resistive to the reactance value and this
411
00:56:42,150 --> 00:56:49,150
stable shows that as the as the value of the
impedance connected in the neutral of the
412
00:56:52,650 --> 00:56:57,650
receiving end side and in the sending end
side are varied from ohmic value to the reactance
413
00:56:57,650 --> 00:57:04,650
value, the stability limit varies thus we
can say that is grounding affects the stability.
414
00:57:05,500 --> 00:57:11,750
Now I can say conclude my presentation that
today we have examined the affect of fault
415
00:57:11,750 --> 00:57:18,750
duration type of fault location of fault and
the grounding on the stability of the system,
416
00:57:24,880 --> 00:57:31,880
thank you.